MathLabs

Problem 4

Let f(θ)=1−acos⁡θ−bsin⁡θ−Acos⁡2θ−Bsin⁡2θf(\theta)=1-a\cos\theta-b\sin\theta-A\cos2\theta-B\sin2\theta, where a,b,A,Ba,b,A,B are real. Prove that if f(θ)≥0f(\theta)\ge0 for all real θ\theta, then a2+b2≤2a^2+b^2\le2 and A2+B2≤1A^2+B^2\le1.
Step 1 of 3: Isolate the second harmonic
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f(θ)+f(θ+π)=2−2Acos⁡2θ−2Bsin⁡2θ≥0f(\theta)+f(\theta+\pi)=2-2A\cos2\theta-2B\sin2\theta\ge0
Detailed analysis

The terms involving a,ba,b cancel under a shift by π\pi. Hence Acos⁡2θ+Bsin⁡2θ≤1A\cos2\theta+B\sin2\theta\le1 for every θ\theta. Its maximum amplitude is A2+B2\sqrt{A^2+B^2}, giving A2+B2≤1A^2+B^2\le1.